The hyperbola

A hyperbola has two asymptotes. An asymptote is a line that a curve approaches as it tends towards infinity but never touches or crosses. The two branches of the hyperbola approach these asymptotes.

To find the asymptotes of a hyperbola, consider what happens as $x$ approaches infinity.

From $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we can rearrange to make $y$ the subject:

$\frac{y^2}{b^2} = \frac{x^2}{a^2} - 1$

$y^2 = b^2\left(\frac{x^2}{a^2} - 1\right)$

$y = \pm \sqrt{b^2\left(\frac{x^2}{a^2} - 1\right)}$

$y = \pm \frac{b}{a}\sqrt{x^2 - a^2}$

$y = \pm \frac{bx}{a}\sqrt{1 - \frac{a^2}{x^2}} \quad \text{(Equation 1.18)}$

From Equation 1.18, as $x \to \infty$, $\frac{a^2}{x^2} \to 0$, which implies:

$y = \pm \frac{bx}{a} \quad \text{(Equation 1.19)}$

Equations (1.19) represent the equations of the asymptotes to the hyperbola.

For a hyperbola of the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$, the asymptotes are given by $y = \pm \frac{b}{a}x$.

Example 1

Find the asymptotes of the hyperbola $\frac{x^2}{3} - \frac{y^2}{4} = 1$.

Solution:

Given $\frac{x^2}{3} - \frac{y^2}{4} = 1$, we have $a^2 = 3$ and $b^2 = 4$, so $a = \sqrt{3}$ and $b = 2$.

The equations of the asymptotes are $y = \pm \frac{b}{a}x = \pm \frac{2}{\sqrt{3}}x = \pm \frac{2\sqrt{3}}{3}x$.

Example 2

Find the equation of the hyperbola in the form $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ whose asymptotes are $y = \pm \frac{12}{13}x$.

Solution:

The asymptotes of the hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ are $y = \pm \frac{b}{a}x$.

Comparing with the given asymptotes $y = \pm \frac{12}{13}x$, we have $\frac{b}{a} = \frac{12}{13}$.

We are only given the ratio of b and a, we cannot find unique values for a and b. However, if we assume a simple case where $a=13$ and $b=12$, then the equation of the hyperbola is $\frac{y^2}{144} - \frac{x^2}{169} = 1$.

Note: There are infinitely many hyperbolas with these asymptotes, as $a$ and $b$ could be any values as long as the ratio $\frac{b}{a}$ is $\frac{12}{13}$.

Similar to the ellipse, the graph of the hyperbola can also be translated. Translating a hyperbola $h$ units horizontally and $k$ units vertically shifts the center to $(h, k)$. This transforms the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ by replacing $x$ with $(x - h)$ and $y$ with $(y - k)$. Therefore, the standard equation of the translated hyperbola with center $(h, k)$ is:

$\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$

Example 1

A hyperbola has its center at $(2, 1)$, one focus at $(-3, 1)$, and the length of the transverse axis is 8 units.

1. Find the other focus.
2. Find the equation of the hyperbola and the asymptotes.
3. Sketch the hyperbola. (Insert sketch here)

Solution:

a. The center is the midpoint of the foci. Let the other focus be $(x, y)$. Then:

$\frac{-3 + x}{2} = 2$ and $\frac{1 + y}{2} = 1$

$-3 + x = 4$ and $1 + y = 2$

$x = 7$ and $y = 1$. The other focus is $(7, 1)$.

b. The transverse axis length is $2a = 8$, so $a = 4$ and $a^2 = 16$. The distance from the center to a focus is $c$. The distance between the x-coordinates of the center and focus gives us c: $|-3-2| = 5$, thus $c = 5$. Since $c^2 = a^2 + b^2$, we have $25 = 16 + b^2$, so $b^2 = 9$ and $b = 3$.

The equation of the hyperbola is $\frac{(x - 2)^2}{16} - \frac{(y - 1)^2}{9} = 1$.

The asymptotes are given by $y - k = \pm\frac{b}{a}(x - h)$, so $y - 1 = \pm\frac{3}{4}(x - 2)$, which gives $y = \frac{3}{4}x - \frac{3}{2} + 1$ and $y = -\frac{3}{4}x + \frac{3}{2} + 1$. Simplifying, we get $y = \frac{3}{4}x - \frac{1}{2}$ and $y = -\frac{3}{4}x + \frac{5}{2}$.

Example 2

Given the hyperbola $9y^2 - 16x^2 - 36y - 32x - 124 = 0$.

1. Express the equation in standard form.
2. Find its center, foci, directrices, vertices, and asymptotes.
3. Sketch its graph.

Solution:

(a) Completing the square:

$9(y^2 - 4y) - 16(x^2 + 2x) = 124$

$9(y^2 - 4y + 4) - 16(x^2 + 2x + 1) = 124 + 36 - 16$

$9(y - 2)^2 - 16(x + 1)^2 = 144$

$\frac{(y - 2)^2}{16} - \frac{(x + 1)^2}{9} = 1$

(b) Comparing with $\frac{(y - k)^2}{b^2} - \frac{(x - h)^2}{a^2} = 1$, we have $h = -1$, $k = 2$, $a = 3$, and $b = 4$. The center is $(-1, 2)$.

$c^2 = a^2 + b^2 = 9 + 16 = 25$, so $c = 5$. The foci are $(-1, 2 \pm 5)$, which are $(-1, 7)$ and $(-1, -3)$.

The vertices are $(-1, 2 \pm 4)$, which are $(-1, 6)$ and $(-1, -2)$.

$e = \frac{c}{b} = \frac{5}{4}$. The directrices are $y = 2 \pm \frac{4}{\frac{5}{4}} = 2 \pm \frac{16}{5}$, which are $y = \frac{26}{5}$ and $y = -\frac{6}{5}$.

The asymptotes are $y - 2 = \pm\frac{4}{3}(x + 1)$, which simplifies to $y = \frac{4}{3}x + \frac{14}{3}$ and $y = -\frac{4}{3}x + \frac{2}{3}$.

Example 3

Find the equation of a hyperbola for which the difference of the distances from the points $A(4, 0)$ and $B(-4, 0)$ is always equal to 2.

Solution:

Let $P(x, y)$ be a point on the hyperbola. Then $|PA - PB| = 2$.

$|\sqrt{(x - 4)^2 + y^2} - \sqrt{(x + 4)^2 + y^2}| = 2$

This is difficult to simplify directly. The foci are at $(\pm 4,0)$ so $c=4$. The difference of distances is $2a=2$, so $a=1$.

Since $c^2=a^2+b^2$ then $16=1+b^2$ so $b^2=15$

The equation is $\frac{x^2}{1} - \frac{y^2}{15} = 1$ or $x^2 - \frac{y^2}{15} = 1$

The equations of the tangent and normal to a hyperbola are derived similarly to other conics. Differentiating the hyperbola's equation yields the tangent's gradient at the point of tangency.

Let $P(x_1, y_1)$ be the point of tangency. The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is derived as follows:

From $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we have $b^2x^2 - a^2y^2 = a^2b^2$.

Implicit differentiation with respect to $x$:

$2b^2x - 2a^2y\frac{dy}{dx} = 0$

$\frac{dy}{dx} = \frac{b^2x}{a^2y}$

The gradient of the hyperbola (and thus the tangent) at $P(x_1, y_1)$ is $m = \frac{b^2x_1}{a^2y_1}$.

The equation of the tangent at $P(x_1, y_1)$ is:

$y - y_1 = \frac{b^2x_1}{a^2y_1}(x - x_1)$

Simplifying (using the fact that $\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$), we get:

$\frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1$

The normal's equation uses the fact that the product of the tangent's and normal's gradients is -1. If $m_1$ is the tangent's gradient and $m_2$ is the normal's gradient, then $m_1m_2 = -1$.

The normal's gradient at $P(x_1, y_1)$ is $m_2 = -\frac{a^2y_1}{b^2x_1}$. Therefore, the equation of the normal is:

$y - y_1 = -\frac{a^2y_1}{b^2x_1}(x - x_1)$

which can be rearranged to $\frac{a^2(x-x_1)}{x_1} + \frac{b^2(y-y_1)}{y_1} = 0$

Example 1

Find the equations of the tangent and normal to the hyperbola $\frac{x^2}{25} - \frac{y^2}{4} = 1$ at the point $(5\sqrt{2}, 2)$.

Solution:

Here, $a^2 = 25$, $b^2 = 4$, $x_1 = 5\sqrt{2}$, and $y_1 = 2$.

Tangent: $\frac{(5\sqrt{2})x}{25} - \frac{2y}{4} = 1 \Rightarrow \frac{\sqrt{2}x}{5} - \frac{y}{2} = 1 \Rightarrow 2\sqrt{2}x - 5y = 10$

Normal: $y-2 = -\frac{25(2)}{4(5\sqrt{2})}(x-5\sqrt{2}) \Rightarrow y-2 = -\frac{5\sqrt{2}}{4}(x-5\sqrt{2}) \Rightarrow 4y - 8 = -5\sqrt{2}x + 50 \Rightarrow 5\sqrt{2}x + 4y - 58 = 0$

Example 2

Find the equations of the tangent and normal to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ at the point $(6, \frac{12}{5})$.

Solution:

Here, $a^2 = 25$, $b^2 = 16$, $x_1 = 6$, and $y_1 = \frac{12}{5}$.

Tangent: $\frac{6x}{25} - \frac{(\frac{12}{5})y}{16} = 1 \Rightarrow \frac{6x}{25} - \frac{3y}{20} = 1 \Rightarrow 24x - 15y = 100$

Normal: $y-\frac{12}{5} = -\frac{25(\frac{12}{5})}{16(6)}(x-6) \Rightarrow y-\frac{12}{5} = -\frac{5}{8}(x-6) \Rightarrow 40y-96=-25x+150 \Rightarrow 25x+40y-246=0$

In Figure 1.14, the line segment $LL'$ represents the latus rectum of a hyperbola. (Insert Figure 1.14 here)

Given the focus $F(ae, 0)$ and the points $L(ae, y)$ and $L'(ae, -y)$. These points satisfy the hyperbola's equation:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Substituting $x = ae$:

$\frac{(ae)^2}{a^2} - \frac{y^2}{b^2} = 1$

$e^2 - \frac{y^2}{b^2} = 1$

$\frac{y^2}{b^2} = e^2 - 1$

$y^2 = b^2(e^2 - 1)$

$y = \pm b\sqrt{e^2 - 1}$

The length of the latus rectum is the distance between $L$ and $L'$:

$LL' = 2b\sqrt{e^2 - 1}$

Since $b^2 = a^2(e^2 - 1)$, we have $\sqrt{e^2 - 1} = \frac{b}{a}$.

Substituting this into the latus rectum equation:

$LL' = 2b\left(\frac{b}{a}\right) = \frac{2b^2}{a}$

Therefore, the length of the latus rectum is $\frac{2b^2}{a}$ units when the transverse axis is horizontal. When the transverse axis is vertical (equation $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$), the length of the latus rectum is $\frac{2a^2}{b}$ units.

Example 1

Given that the lengths of the latus rectum and transverse axis of the hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ are $\frac{13}{2}$ and 16 units, respectively. Find the equation of the hyperbola.

Solution:

The latus rectum length is $\frac{2a^2}{b} = \frac{13}{2}$, so $4a^2 = 13b$ (Equation i).

The transverse axis length is $2b = 16$, so $b = 8$ (Equation ii).

Substitute Equation ii into Equation i:

$4a^2 = 13(8)$

$4a^2 = 104$

$a^2 = 26$

The equation of the hyperbola is $\frac{y^2}{64} - \frac{x^2}{26} = 1$.

Example 2

Find the length of the latus rectum of the hyperbola $\frac{(x + 1)^2}{9} - \frac{(y - 2)^2}{4} = 1$.

Solution:

Given $\frac{(x + 1)^2}{9} - \frac{(y - 2)^2}{4} = 1$, we have $a^2 = 9$ and $b^2 = 4$, so $a = 3$ and $b = 2$.

The length of the latus rectum is $\frac{2b^2}{a} = \frac{2(4)}{3} = \frac{8}{3}$ units.

Similar to parabolas and ellipses, hyperbolas can be represented in parametric form. Parametrization expresses an equation with multiple variables in terms of a single variable (a parameter). Consider the hyperbola in Figure 1.15. The circle is the auxiliary circle with radius $a$ (half the transverse axis).

The line segment $OM$ is a radius of the auxiliary circle ($OM = a$). $MN$ is tangent to this circle, and angle $\theta$ is the eccentric angle.

In right-angled triangle $\triangle OMN$:

$\cos\theta = \frac{OM}{ON} = \frac{a}{x}$

$x = a\sec\theta$

Recall the hyperbola's equation: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Substitute $x = a\sec\theta$:

$\frac{(a\sec\theta)^2}{a^2} - \frac{y^2}{b^2} = 1$

$\sec^2\theta - \frac{y^2}{b^2} = 1$

$\frac{y^2}{b^2} = \sec^2\theta - 1$

$y^2 = b^2(\sec^2\theta - 1)$

Since $\sec^2\theta - 1 = \tan^2\theta$:

$y^2 = b^2\tan^2\theta$

$y = \pm b\tan\theta$

Therefore, $x = a\sec\theta$ and $y = b\tan\theta$ are the parametric equations when the transverse axis is along the x-axis. The point $P(x, y)$ can be represented as $P(a\sec\theta, b\tan\theta)$.

If the transverse axis is along the y-axis, the parametric equations are $x = a\tan\theta$ and $y = b\sec\theta$.

For a translated hyperbola with center $(h, k)$, the parametric equations are $x = h + a\sec\theta$ and $y = k + b\tan\theta$ (for a horizontal transverse axis) or $x = h + a\tan\theta$ and $y = k + b\sec\theta$ (for a vertical transverse axis).

Example 1

Express the hyperbola $\frac{x^2}{16} - \frac{y^2}{36} = 1$ in parametric form.

Solution:

Here, $a^2 = 16$ and $b^2 = 36$, so $a = 4$ and $b = 6$.

Since the transverse axis is horizontal, the parametric equations are $x = 4\sec\theta$ and $y = 6\tan\theta$.

Example 2

Write the parametric equations of the hyperbola $\frac{(x - 2)^2}{16} - \frac{(y - 4)^2}{9} = 1$.

Solution:

This is a translated hyperbola with center $(h, k) = (2, 4)$, $a^2 = 16$ (so $a = 4$), and $b^2 = 9$ (so $b = 3$).

Since the transverse axis is horizontal, the parametric equations are $x = 2 + 4\sec\theta$ and $y = 4 + 3\tan\theta$.